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Senior Manager  S
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According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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Question Stats: 80% (01:35) correct 20% (02:22) wrong based on 1514 sessions

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According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

a) 5:30
b) 7:00
c) 7:30
d) 8:00
e) 9:00

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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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Baten80 wrote:
According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)^²+500 for 0≤t≤10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

a) 5:30
b) 7:00
c) 7:30
d) 8:00
e) 9:00

Don't get bogged down by the dirty N(t) expression. Just think of it this way:

N(t) is a combination of two terms: a positive term (500) and a negative term ($$-20(t-5)^2$$).
To maximize N(t), I need to make the positive term as large as possible (It is a constant here so I cannot do much with it) and the absolute value of the negative term as small as possible. The smallest absolute value is 0. Can I make it 0? Yes, if I make t = 5, the negative term becomes 0 and N(t) is maximized. My answer must be 2:00 + 5 hrs i.e. 7:00.

Most of the maximum minimum questions on GMAT will require you to only think logically. The calculations involved will be minimum.
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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2
the expression is N(t)= -20(t-5)^²+500
(of course valid after 2:00 in the morning)

"the depth would be maximum" means the value of the above expression should be maximum
or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero

the square part is zero at t=5

so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B)
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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puneetkr wrote:
the expression is N(t)= -20(t-5)^²+500
(of course valid after 2:00 in the morning)

"the depth would be maximum" means the value of the above expression should be maximum
or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero

the square part is zero at t=5

so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B)

if -20(t-5)^²=0
then t = 5
But Why is the 500 of the equation is not considered? Show your complete calculation.
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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this is the complete calculation

N(t)= -20(t-5)^²+500
now for any value of t we put in; we get some negative value of -20(t-5)^² (say -x)
so our expression is now N(t)=500-x

this expression would have maximum value only when x is minimum
we know the minimum value for a square term is "zero" and (x has a square term)
and that comes when t=5
i.e. when we put t=5 here we get N(t) = 500-0 = 500
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Joined: 02 Sep 2009
Posts: 61189
Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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1
Baten80 wrote:
puneetkr wrote:
the expression is N(t)= -20(t-5)^²+500
(of course valid after 2:00 in the morning)

"the depth would be maximum" means the value of the above expression should be maximum
or the value of square term (which has a negative 20 attached to it) should be minimum i.e. zero

the square part is zero at t=5

so the time at which the depth is maximum is 2:00 + 5 hrs= 7:00 (B)

if -20(t-5)^²=0
then t = 5
But Why is the 500 of the equation is not considered? Show your complete calculation.

According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)^²+500 for 0≤t≤10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

A. 5:30
B. 7:00
C. 7:30
D. 8:00
E. 9:00

Consider this: $$-20(t-5)^2\leq{0}$$ hence $$500-20(t-5)^2$$ reaches its maximum when $$-20(t-5)^2=0$$, thus when $$t=5$$. 2:00AM+5 hours=7:00AM.

Hope it helps.
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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Intern  Joined: 07 Dec 2012
Posts: 9
Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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Because if I plug in t = 3.5 (which is 5.30 (3.5 after the 2:00) , it gives me the maximum value of -20(t-5)^2 expression, which will be positive. Hence, the height will be at its max of 545. Hence, shouldn't the answer be A?

Thank you
Math Expert V
Joined: 02 Sep 2009
Posts: 61189
Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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herein wrote:

Because if I plug in t = 3.5 (which is 5.30 (3.5 after the 2:00) , it gives me the maximum value of -20(t-5)^2 expression, which will be positive. Hence, the height will be at its max of 545. Hence, shouldn't the answer be A?

Thank you

Also, if t = 3.5, then 500 - 20(3.5 - 5)^2 = 455, not 545.
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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1
Baten80 wrote:
According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?0

N(t)= $$-20(t - 5)^2$$ + 500 ;

The red highlighted portion of the equation is the most critical part, further t = 2am + time past 2am

Now why $$-20(t - 5)^2$$ is important ;

Any value of t less than 5 will result in a negative value of $$-20(t - 5)^2$$

Lets check :

Time : 3am , t = 1 { t = 2am + time past 2am }

$$-20(t - 5)^2$$

$$-20(1 - 5)^2$$

$$-20(-4)^2$$

$$-20(16)$$

$$-320$$

When put in the final equation : $$-20(t - 5)^2$$ the result will be -320 + 500 => 180

Check a few more any value of t less than 5 will result in -ve value of $$-20(t - 5)^2$$ and will ultimately lead to a depth of water less than 500.

Our target is getting t = 5 ; since at t = 5

$$-20(t - 5)^2$$ will be 0 and the final equation $$-20(t - 5)^2$$ + 500 will be maximum ; ie 500

We already know t = 2am + time past 2am

So, 7 = 2am + 5 hours.

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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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3
2
Hi All,

These specific types of "limit" questions are relatively rare on Test Day, although you'll likely be tested on the concept at least once. Whenever you're asked to minimize or maximize a value, you should look to do something with the other "pieces" of the equation (usually involving maximizing or minimizing those pieces).

In the given equation, notice how you have two "parts": the -20(something) and a +500. Here, to MAXIMIZE the value of N(t), we have to minimize the "impact" that the -20(something) has on the +500. By making that first part equal 0, we'll be left with 0 + 500. Mathematically, we have to make whatever is inside the parentheses equal 0....

(T-5) = 0

T = 5

Since T represents the number of hours past 2:00am, we know that at 7:00am, the water will reach 500cm (the maximum value).

GMAT assassins aren't born, they're made,
Rich
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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1
A useful trick to solve Maximum of any equation,with a single vraiable:

Just differentiate the equation with the variable and equate it to zero to find the solution. That solution is the value at which the euqation will have optimum value.

N(t)= -20(t-5)^²+500

d/dt N(t)= -40(t-5)=0

-40t=-200 => t=5

So, 7 = 2am + 5 hours.
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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valardohaeris wrote:
A useful trick to solve Maximum of any equation,with a single vraiable:

Just differentiate the equation with the variable and equate it to zero to find the solution. That solution is the value at which the euqation will have optimum value.

N(t)= -20(t-5)^²+500

d/dt N(t)= -40(t-5)=0

-40t=-200 => t=5

So, 7 = 2am + 5 hours.

Hi,
I used that method too.

I would like to know if there are always questions like this on every exam.

BR
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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Baten80 wrote:
According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

a) 5:30
b) 7:00
c) 7:30
d) 8:00
e) 9:00

Since -20(t - 5)^2, will be a nonpositive number, its maximum value is 0 when t = 5, and the maximum value of the function will then be:

N(5) = -20(5 - 5)^2 + 500 = 500

Thus, the maximum depth is at 2am + 5 hours = 7am.

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According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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Can someone explain why / how you would know not to simplify the original provided equation (-20(t-5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:

-20(t-5)^2 + 500
-20(t^2 - 10t + 25) + 500 <-- foil
-20t^2 + 200t - 500 + 500 <-- distributed the -20
-20t^2 + 200t
20t(t + 10t) <-- simplified formula

Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past.
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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rjacobson99 wrote:
Can someone explain why / how you would know not to simplify the original provided equation (-20(t-5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:

-20(t-5)^2 + 500
-20(t^2 - 10t + 25) + 500 <-- foil
-20t^2 + 200t - 500 + 500 <-- distributed the -20
-20t^2 + 200t
20t(t + 10t) <-- simplified formula

Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past.

Hi rjacobson99,

In the last "step" of you work, you have not properly accounted for the 'minus' sign. If you want to factor out "20t", then that's fine, but here's what you would be left with:

-20t^2 + 200t
(20t)(-t + 10)
(20t)(10 - t)

The maximum result will occur when t = 5.

GMAT assassins aren't born, they're made,
Rich
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According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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EMPOWERgmatRichC wrote:
rjacobson99 wrote:
Can someone explain why / how you would know not to simplify the original provided equation (-20(t-5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:

-20(t-5)^2 + 500
-20(t^2 - 10t + 25) + 500 <-- foil
-20t^2 + 200t - 500 + 500 <-- distributed the -20
-20t^2 + 200t
20t(t + 10t) <-- simplified formula

Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past.

Hi rjacobson99,

In the last "step" of you work, you have not properly accounted for the 'minus' sign. If you want to factor out "20t", then that's fine, but here's what you would be left with:

-20t^2 + 200t
(20t)(-t + 10)
(20t)(10 - t)

The maximum result will occur when t = 5.

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

In the correct version of the formula (20t)(10-t), is there a way to determine the inflection point of t = 5 without having to plug in values from t = 1 to 6? I.e. is there a shortcut?

RJ
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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rjacobson99 wrote:
EMPOWERgmatRichC wrote:
rjacobson99 wrote:
Can someone explain why / how you would know not to simplify the original provided equation (-20(t-5)^2 + 500)? When you simplify the equation, you seems to get a different answer. Please see work below:

-20(t-5)^2 + 500
-20(t^2 - 10t + 25) + 500 <-- foil
-20t^2 + 200t - 500 + 500 <-- distributed the -20
-20t^2 + 200t
20t(t + 10t) <-- simplified formula

Given the 20t(t+10t) is simplified correctly, then the water level will continue to grow with every hour with the latest hour being the maximum. This is different than the answer provided, where t is at the max once 5 hours past.

Hi rjacobson99,

In the last "step" of you work, you have not properly accounted for the 'minus' sign. If you want to factor out "20t", then that's fine, but here's what you would be left with:

-20t^2 + 200t
(20t)(-t + 10)
(20t)(10 - t)

The maximum result will occur when t = 5.

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

In the correct version of the formula (20t)(10-t), is there a way to determine the inflection point of t = 5 without having to plug in values from t = 1 to 6? I.e. is there a shortcut?

RJ

Hi rjacobson99,

Unfortunately, manipulating the equation in the way that you did places the variable "t" in both pairs of parentheses, so there isn't an 'obvious' solution to maximize the value. However, in the original equation, there IS an obvious Number Property that you can use...

N(t) = -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10.

In the given equation, notice how you have two "parts": the -20(something) and a +500. Here, to MAXIMIZE the value of N(t), we have to minimize the "impact" that the negative term - the -20(something) - has on the +500. By making that first part equal 0, we'll be left with 0 + 500. Mathematically, we have to make whatever is inside the parentheses equal 0....

(T-5) = 0

T = 5

GMAT assassins aren't born, they're made,
Rich
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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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This is beyond the scope of what's tested on the GMAT, but using derivatives, we can answer this in 15 seconds. The derivative will always be where the quadratic is at its maximum.

Given: $$N(t) = -20 (t-5)^2 + 500$$
Take the derivative: (t-5) = 0
Solve: t = 5
Plug into prompt: 2:00 am + 5 hours = 7:00 am

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Re: According to a certain estimate, the depth N(t), in centimeters, of  [#permalink]

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Baten80 wrote:
According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

a) 5:30
b) 7:00
c) 7:30
d) 8:00
e) 9:00

According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t - 5)^2 + 500 for 0 ≤ t ≤ 10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

For t = 5 ; N(t) = 500 max
5 hours past 2:00 = 7:00

IMO B Re: According to a certain estimate, the depth N(t), in centimeters, of   [#permalink] 16 Sep 2019, 07:40
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